Question:
Does the point $(-2.5,3.5)$ lie inside, outside or on the circle $x^{2}+y^{2}=25 ?$
Solution:
The equation of the given circle is $x^{2}+y^{2}=25$.
$x^{2}+y^{2}=25$
$\Rightarrow(x-0)^{2}+(y-0)^{2}=5^{2}$, which is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $h=0, k=0$, and $r=5$
$\therefore$ Centre $=(0,0)$ and radius $=5$
Distance between point $(-2.5,3.5)$ and centre $(0,0)$
$=\sqrt{(-2.5-0)^{2}+(3.5-0)^{2}}$
$=\sqrt{6.25+12.25}$
$=\sqrt{18.5}$
$=4.3$ (approx.) $<5$
Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.